The factorial 4! is equal to
\[4!=4*3*2*1\]
24?
Yes!
good stuff
Or just use a calculator that has the factorial sign
i tried finding the factorial of LGBAAA!!! in my scientific calculator. said "syntax error"
\[\large n! = n \times (n-1) \times (n-2) \times (n-3) \times .............. 3 \times 2 \times 1\] Put n= 4 here..
@waterineyes n! = 4 * 3 * 2 * 1 * 3 * 2 * 1 = 144 O:
lol should only be \[n! = n \times (n-1) \times (n-2) \times \cdots \times 1\]
its the same thing igba
Unrolling factorials.
nope... x 3 x 2 x 1 is different like @nphuongsun93 said
Really??? Well, I got: \[4! = 4 \times (4-1) \times (4-2) \times (4-3) = 4 \times 3 \times 2 \times 1 = 24\]
where's x 3 x 2 x 1 there? lol
What waterineyes wrote is the same as what Igba wrote; but igba just wrote it in a shorter form. From memory you can also wrote it as: \(n! = n(n-1)!\)
@waterineyes i'm just teasing you we all know this
That in the decreasing sense I wrote there... You are also right lgba.. if we put 1 in the last then this will reduce confusion..
but the dotdotdot does express something ;)
No @Mimi_x3 you can also right it as: \[n! = n(n-1)(n-2)!\] Ha ha ha..
No you are wrong lgba.. Careful..
?
\[n! = (n(n-n)!)!\]
Wel, from my notes \(n! = (n-1)!*n\) Mynotes cannot be wrong.
@Mimi_x3 you can subtract 1 upto the length you want..
\[n! = (n \times 1)!\]
\[n! = (n \times 0!)!\]
\[ \huge \huge n! = n! \]
\[n! = 42!\]
\[\large n! = \sqrt{(n!)^2}\]
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